Answer:
6.10%
Step-by-step explanation:
For every round since we are choosing with equal probability, then the probability of choosing from either of the bags is 1/2
Here, this scenario is only possible when a particular bag has been chosen 10 times and the other bag has been chosen 6 times ( meaning a particular bag has been emptied and for a particular bag to be emptied, all of its content would have been picked)
Now on the 17th trial, an empty bag is chosen
Therefore the required probability will be;
16C0 * (1/2)^10 * (1/2)^6 * 1/2 = 0.0610 = 6.1%
Divide both sides by <span>t
</span>
<span>a+b=r/t
</span>Subtract b <span>from both sides
</span>
a<span>=r/t-b = Solution </span>
Answer:
depended is the 6th grade
independed the 2/9 students
-by-step explanation:
Answer:
Step-by-step explanation:
Given data:
SS={0,1,2,3,4}
Let probability of moving to the right be = P
Then probability of moving to the left is =1-P
The transition probability matrix is:
![\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26P%260%260%260%5C%5C1-P%261%26P%260%260%5C%5C0%261-P%261%26P%260%5C%5C0%260%261-P%261%26P%5C%5C0%260%260%261-P%261%5Cend%7Barray%7D%5Cright%5D)
Calculating the limiting probabilities:
π0=π0+Pπ1 eq(1)
π1=(1-P)π0+π1+Pπ2 eq(2)
π2=(1-P)π1+π2+Pπ3 eq(3)
π3=(1-P)π2+π3+Pπ4 eq(4)
π4=(1-P)π3+π4 eq(5)
π0+π1+π2+π3+π4=1
π0-π0-Pπ1=0
→π1 = 0
substituting value of π1 in eq(2)
(1-P)π0+Pπ2=0
from
π2=(1-P)π1+π2+Pπ3
we get
(1-P)π1+Pπ3 = 0
from
π3=(1-P)π2+π3+Pπ4
we get
(1-P)π2+Pπ4 =0
from π4=(1-P)π3+π4
→π3=0
substituting values of π1 and π3 in eq(3)
→π2=0
Now
π0+π1+π2+π3+π4=0
π0+π4=1
π0=0.5
π4=0.5
So limiting probabilities are {0.5,0,0,0,0.5}
' x ÷ 3 ' is just another way to write ' 1/3 x ' ,
so your equation is y = 1/3 x .
The graph of that equation is a straight line,
going through the origin, with a slope of 1/3 .